Question

A box contains $6$ red, $5$ blue and $4$ white marbles. Four marbles are chosen at random without replacement. The probability that there is atleast one marble of each colour among the four chosen, is?

Answer 1

Probability of selecting a blue ball in $1^{st}$ attempt = $\frac{5}{15}$

Probability of selecting a red ball in $2^{nd}$ attempt = $\frac{6}{14}$

Probability of selecting a white ball in $3^{rd}$ attempt = $\frac{4}{13}$

They can re-arrange themselves in $\frac{4!}{2!}$ ways.

Probability = $\frac{4!}{2!}\times \frac{5}{15}\times \frac{6}{14}\times \frac{4}{13}=\frac{48}{91}$